Problem: Suppose that all four of the numbers \[3 - 2\sqrt{2}, \; -3-2\sqrt{2}, \; 1+\sqrt{7}, \; 1-\sqrt{7}\]are roots of the same nonzero polynomial with rational coefficients. What is the smallest possible degree of the polynomial?
Answer: Because the polynomial has rational coefficients, the radical conjugate of each of the given roots must also be roots of the polynomial. However, $1+\sqrt{7}$ and $1-\sqrt{7}$ are each other's radical conjugates, so we only get $2$ more roots. (You might be tempted to think that $3-2\sqrt2$ and $-3-2\sqrt2$ are also a pair of radical conjugates, but the radical conjugate of $3-2\sqrt2$ is $3+2\sqrt2,$ while the radical conjugate of $-3-2\sqrt2$ is $-3+2\sqrt2.$ Therefore, each one of the numbers $3-2\sqrt2$ and $-3-2\sqrt2$ is actually the negation of the radical conjugate of the other one.) In total, the polynomial must have at least $4+2=6$ roots.

Furthermore, the polynomial
\[(x - 3 + 2 \sqrt{2})(x - 3 - 2 \sqrt{2})(x + 3 + 2 \sqrt{2})(x + 3 - 2 \sqrt{2})(x - 1 - \sqrt{7})(x - 1 + \sqrt{7}) = (x^2 - 6x + 1)(x^2 + 6x + 1)(x^2 - 2x - 6)\]has roots $3 \pm 2 \sqrt{2},$ $-3 \pm 2 \sqrt{2},$ and $1 \pm \sqrt{7},$ and has rational coefficients.  Hence, the smallest possible degree is $\boxed{6}.$